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# Find the value of k for which the following pair of linear equations have infinitely many solutions.

$2x+3y=7;\ ( k-1) x+( k+2) y=3k$

$2x+3y=7;\ ( k-1) x+( k+2) y=3k$

**Given:**A pair of linear equations $2x+3y=7;\ ( k-1) x+( k+2) y=3k$

**To do:**To find out the value of k for which this pair of equations have infinitely many solutions.

**Solution:**

**If there are two equations $a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$**

For the infinitely many solutions there is a condition

$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} \ $

Here given linear equations:

$2x+3y=7$

$( k-1) x+( k+2) y=3k\ $

Here $a_{1} =2$,$b_{1} =3$ and $c_{1} =7$

$a_{2} =( k-1)$ ,$b_{2} =( k+2)$ and $c_{2} =3k$

For infinitely many solutions for this pair of equations it should be.

$\frac{2}{k-1} =\frac{3}{k+2} =\frac{7}{3k}$

$\frac{2}{k-1} =\frac{3}{k+2}$

$\Rightarrow 2( k+2) \ =3( k-1)$

$\Rightarrow 2k+4\ =\ 3k-3$

$\Rightarrow 3k-2k=4+3$

$\Rightarrow k=7$

**Hence for the value of $k=7$ this pair of equations will have infinitely many solutions.**

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